$c) n_{C_2H_2}=n_{Ag_2C_2}=0,135(mol)$

$1<\dfrac{n_{Br_2}}{n_{C_2H_2}}=1,33<2 \rightarrow$ thu dc 2 sp 

$C_2H_2+Br_2\rightarrow C_2H_2Br_2$

$C_2H_2+2Br_2\rightarrow C_2H_2Br_4$

$n_{C_2H_2Br_4}=n_{Br_2}-n_{C_2H_2}=0,045(mol);n_{C_2H_2Br_2}=0,135-0,045=0,09(mol)$

$\rightarrow m_{C_2H_2Br_4}=0,045.346=15,57g;m_{C_2H_2Br_2}=0,09.186=16,74g$

`a)`

TN1:

`C_2H_4+Br_2->C_2H_4Br_2`

`C_2H_2+2Br_2->C_2H_2Br_4`

TN2:

`C_2H_2+2AgNO_3+2NH_3->C_2Ag_2+2NH_4NO_3`

`b)`

`X:C_3H_8:x(mol);C_2H_4:y(mol);C_2H_2:z(mol)`

`->x+y+z={5,57775}/{24,79}=0,225(1)`

`x=n_{khí\ thoát}={1,3944375}/{24,79}=0,05625(2)`

`z=n_{C_2Ag_2}={32,4}/{240}=0,135(3)`

`(1)(2)(3)->y=0,03375(mol)`

`\%V_{C_3H_8}={0,05625}/{0,225}.100\%=25\%`

`\%V_{C_2H_4}={0,03375}/{0,225}.100\%=15\%`

`\%V_{C_2H_2}=100-15-25=60\%`

`c)`

`C_2H_2+Br_2->C_2H_2Br_2`

`0,135->0,135->0,135(mol)`

`n_{Br_2\ du}=0,18-0,135=0,045(mol)`

`C_2H_2Br_2+Br_2->C_2H_2Br_4`

`0,045←0,045->0,045(mol)`

`m_{C_2H_2Br_4}=0,045.346=15,57(g)`

`m_{C_2H_2Br_2}=(0,135-0,045).186=16,74(g)`